# Particle Creation by a Classical Source (on-shell mass momenta)

causalityfourier transformoperatorsquantum-field-theoryspecial-relativity

It is noted in Peskin and Schroeder's QFT text that the momenta used in the evaluation of the field operator $$\phi(x)$$ are "on mass-shell": $$p^2=m^2$$. Specifically, this is in relation to the expression:
$$\phi(x)=\int{d^3p\over (2\pi)^3}{1\over\sqrt{2E_p}}\bigg\{\bigg(\mathbf{a_p}+{i\over\sqrt{2E_p}}\tilde{j}(p)\bigg)e^{-ip\cdot x}+\;\text{h.c.}\bigg\}\quad \tag{2.64}.$$
Why is this the case? My educated guess is that the particle momenta must be "on shell" because without this condition, $$\phi(x)$$ will not be guaranteed to have Lorentz invariance.

#### Best Answer

TL;DR: The on-shellness of the Fourier decomposition (2.64) comes from the source-full/inhomogeneous Klein-Gordon equation (2.61).

P&S argue this in 2 steps:

1. First P&S consider the Fourier-transformed source-free/homogeneous Klein-Gordon equation, and argue that its solution $$\phi_0$$ is on-shell.

2. Next P&S introduce in eq. (2.63) a classical source $$j$$ via the retarded Greens function, which can also be brought on an on-shell form: \begin{align} D_R(x-y)~=~~~&\int\!\frac{d^4p}{(2\pi)^4}\frac{i}{(p^0+i\epsilon)^2-E_{\bf p}}e^{-ip\cdot(x-y)}\cr ~=~~~&-\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \frac{dp^0}{2\pi i}\left(\frac{1}{p^0+i\epsilon-E_{\bf p}}-\frac{1}{p^0+i\epsilon+E_{\bf p}}\right)e^{-ip\cdot(x-y)}\cr ~\stackrel{\text{Res.Thm.}}{=}&\theta(x^0-y^0)\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \left(\left. e^{-ip\cdot(x-y)}\right|_{p^0=E_{\bf p}} -\left. e^{-ip\cdot(x-y)}\right|_{p^0=-E_{\bf p}} \right)\cr ~=~~~&\theta(x^0-y^0)\int\!\frac{d^3p}{2E_{\bf p}(2\pi)^3} \left.\left( e^{-ip\cdot(x-y)} - e^{ip\cdot(x-y)} \right)\right|_{p^0=E_{\bf p}}, \end{align}\tag{2.58} where the above $$i\epsilon$$ prescription is consistent with the figure on p.30 between eqs. (2.54-55).

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; section 2.4.